Hydrologic Design 4501

Chapter 1: Hydrologic Water Cycle and Mass Balance

Ardeshir Ebtehaj
University of Minnesota
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Table of Contents

1- Introduction

Spread of human population and formation of civilizations are correlated with the availability of fresh water.
The Mesopotamian civilization, widely considered as the cradle of civilization, was in the the area between the Tigris and Euphrates Rivers. Contemporary distribution of human population in the United State is not an exemption. It is easy to see that the climatology of precipitation somewhat has dictated the density of human population in the U.S. Most of the biggest cities and economies of the country are in the vicinity of lakes, rivers, and estuaries.
These days, hydrologic sciences are largely focused on terrestrial water cycle in continental and global scales under natural condition or anthropogenic impacts. Hydrologic and hydraulic engineering invlove the use and trasport of water through human control or intervention.
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In Minnesota, the population is centered around the Twin Cities metropolitan area, which is at the confluence of the Mississippi and Minnesota rivers. Twin Cities are also located in the southeastern portion of the state that receives the most amount rainfall.
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2- Hydrology and Water Resources Managment

For engineers, hydrologic knowledge is applied to the use and control of water resources on the land areas of the Earth, manily for the following purposes:
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Water resources management is an interdisciplinary field that involves branches of biological sciences, engineering, physical science, and social sciences.
Water resource engineering largely involves hydrologic and hydraulic processes for water supply and water excess management as well as environmental restoration. Movement of water is explained through the laws of fluid mechanics. Application of fluid mechanics for civil engineers often manifest itself in explaining hydrologic and hydraulic processes.
Hydraulic processes include three types of water flows: (1) pressurized pipe flows, (2) open channel flows, and (3) ground-water flows.
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3- Global Freshwater Resources

The following table reports water resorviors of the Earth. As is evident, the largest reservoir of freshwater is in glaciers (i.e., 1.74% of total water and 68.7% of global freshwater) and mainly in Antarctic and Greenland ice sheets. As reported, water and lakes and river are less than 0.3% of total freshwater on Earth.
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4- Water Use in the United States

Water use from a hydrologic perspective is defined as all water flows that are results of human intervention in the hydrologic cycle. The national water use information program conducted by the United States Geological Survey (USGS) distinguishes the following water-use flows as follows:
  1. waterwithdrawals for off-stream purposes
  2. water deliveries at point of use or quantities released after use,
  3. consumptive use
  4. conveyance loss
  5. reclaimed wastewater
  6. return flow
  7. in-stream flow.
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The relationships among the human-made flows at various points of measurement are illustrated in the left and and the an estimate of water use in the United States is shown in the right hand side in billion gallons per day [bgd].

5- Global Water Cycle

Hydrologic cycle occurs through water fluxes across different water reservoirs.
Water flux is the mass of water, per unit area, per time across the water reservoirs in the hydrosphere []. If we devide the flux of water by its density the unit will be [], which is the unit of flow dischrage or the volumetirc water flux.The water fluxes can be in various phases such as vapor, liquid, and/or solid.
Depending on the net outflow discharge and volume of these reservoirs , each has a different residence time . Residence time of the main water reservoirs are:
Among these reservoirs, atmosphere shows the minimum residence time, an indication of its fast evolving dynamics. Groundwater residence time needs to be taken into consideration for sustainable developments in arid and semi-arid regions.
The National Aeronautics and Space Administration (NASA) of the United States prepared a clip (link: NASA: Earth's Water Cycle) that shows the global hydrologic water cycle and the role of Earth observaing satellites in measuring its changes.
A conceptual schematic of the hydrologic water cycle and involved processes and water fluxes is shown in the following figure.
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5-1 Hydrologic watersheds

All of the above processes can be quantified over a hydrologic unit called watershed.
Watersheds can be delineated either manually or automatically using digital elevation models (DEM) and computational algorithms. The size of a watershed can range from a few acres to millions of square miles (Mississippi River basin, 3.2 million square miles) -- depending on the geomorphologic characteristics of land surfaces and location of the outlet. Watersheds have a nested structure.
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5-2 Driver of the global hydrologic water cycle

Solar radiative energy from the Sun is the main driver of the global water cycle. Although the earth is a closed thermodynamic system (no mass exchange with its environment), it is not an isolated system as it exchanges energy with the outer space. This exchange is a consequence of the annual average solar radiation flux of 342 , over the entire Earth surfaces, at the top of its atmosphere.
Due to the ellipsoidal shape and orbital geometry of earth, the equators receive more energy than polar regions. This differential energy budget eventually leads to a pressure gradient in the air atmosphere from equators to poles that circulates the Earth's atmosphere. The moving air masses transport water vapor from tropical oceans and precipitate them over lands. The precipitation water returns back to the atmosphere and oceans through the explained fluxes and processes such as evapotranspiration, infiltration, percolation, runoff, and streamflow.
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5-3 Earth's atmospheric circulation

As explained, the convective motion of air and water masses in atmosphere are mainly because of an existing pressure gradient. Why does such a pressure gradient exist?
As a general rule, when the temperature of an air parcel increases at a constant pressure, based on the ideal gas law, its density reduces and vice versa.
Therefore, in general, a cold air mass is denser than a warm air mass at constant pressure. Because the air is compressible, the density of cold air column decays faster than the density of a warm air column from earth surface to higher altitudes. Due to higher pressure of warm air column aloft than the cold air column, the air flows from warm to cold regions at high elevations.
As the air flows from warm to cold areas, the cold air column becomes heavier and its pressure increases at the surface. As a result, a surface air flow forms from cold to warm areas which eventually gives rise to a circulation pattern.
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The circulation of the earth's atmosphere is due to formation of a pressure gradient from warm (tropics) to cold (polar regions) air masses.

According to the above simple conceptualization of the atmospheric circulation, one may think of the Earth atmospheric circulation as shown in the figure below. However, the one-cell circulation model does not exist because these large eddies are unstable and more importantly there are other forces acting on the traveling air parcels that break these large hypothetical circulation cells apart. One of the main forces is the called the Coriolis force.
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In simple terms, with respect to the a rotating reference frame, it appears that the Coriolis force is deflecting an object that moves from the center to the perimeter and vice versa.
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Coriolis force deflects an object thrown from point A to point B to the right of its path in a counter clockwise rotational system. This force is due to a combined effect of the angular () and linear velocity ().
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The Coriolis force deflects air parcels to the right and left of the their moving direction in the Northern and Southern Hemisphere, respectively, as shown in the above figure.
As a result of the coriolis and pressure gradient forces, those large hypothetical circulation cells breaks into three smaller cells as shown below. It is easy to conclude that the surface air flows from mid-latitudes high-pressure divergence zones towards the tropical low-pressure convergence zones are deflected towards east. These surface air flows are called Easterlies or trade winds. On the other hand, surface air flows from high-pressure horse latitudes towards polar fronts are deflected to the west and create the Westerlies.
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The coriolis force has another important implications in formation and direction of tropical and extra tropical storms. In regions within ±5-20 latitudes, the coriolis force is strong enough to create cyclonic activities around the high and low pressure areas. Based on the explained coriolis effect, it is easy to understand that the air flows counter clockwise around lows and clockwise around highs in the Northern Hemisphere.
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A schematic of airflow in a low pressure cell (left) with convergence at the surface and divergence aloft. Airflows in a high pressure cell diverges at the surface and converges aloft. A satellite footage of the Hurricane Irma shows its counter clock-wise rotation around a strong low-pressure system.

6- Water Budget Analysis

Water budget analysis is extremely important for sustainability of socioeconomic growth.
Formerly, one of the four largest lakes in the world with an area of 68,000 sq-km (26,300 sq-mi), the Aral Sea has been steadily shrinking since the 1960s after the rivers that fed it were diverted by Soviet irrigation projects. By 1997, it had declined to 10% of its original size.
The right hand side shows Iran’s Lake Urmia, which was one of the largest saltwater lakes in the world. With eight times as much salt as seawater, it was the globe’s largest habitat for brine shrimps, which attract flamingos, other birds as they migrate across Asia. But the lake is shirinking. Using images captured by USGS-NASA Landsat satellites, it is determined that its area has decreased by 88% since the 1970s. Many blamed severe drought, but climate data from satellites and other sources demonstrated that the lake is shrinking even in wet years. Therefore, research suggests that the reason is unsustainable agricultural developments. As water retreats, it leaves behind a salty crust that is swept into the air by dust storms. These particles can cause respiratory problems in humans and on nearby agricultural lands. The bottom panel shows that the Great Salt Lake in Utah, United States, is also shrinking.
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Water mass balance analysis in hydrologic systems assesses storage of water in natural or man-made reservoirs and evaluates the amount of water flows across those reservoirs. We often use a system approach to explain hydrologic mass balance analyses. Mass of water parcels is considered as a conserved quantity and a simple mass balance over a system of water storage can be cast as follows:

Thus, we have

In a steady state condition, the mass balance reduces to becuase .
Note that we often express hydrologic flows per unit area (volumetric flux) . The rive flow is often represnted in , precipitation in and evapotranspiration in . Therefore, in water budget analysis, it is important to pay attention to the units and thier proper conversions.
Some relevant important units are:
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On a global scale as earth is a close thermodynamic system, water mass balance at the earth's surface is . In a steady state condition, this mass balance leads to on a global scale.
Making a steady state assumption depends on the time-scale. The global steady state assumption is valid for sufficiently long period of time, because evapotranspiration and precipitation are highly dynamic processes that vary significantly in a shorter period of time. Is this assumption valid in seasonal and annual time scales? Observational evidence suggests that the annual rate of both precipitation and ET is approximately 1000 .
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At a watershed scale, each basin may exchange water mass with its surrounding basins through the underlying groundwater systems. A watershed mass balance can be written as follows:
After averaging over sufficiently long period of time, for a steady state condition, we have
Over short period of time (e.g., less than a year), long resident time of groundwater implies :
Important note: When a storm is occurring, the air humidity is relatively high, which reduces the ET significantly. Assuming (), the watershed mass balance can be reduced to
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Example Problem 1.1: Determine the volume of water lost in cubic meters due to evaporation during a year from the surface of a 1500-ha terminal lake located in a region where the annual rainfall is 50 cm. The increase in the depth of the lake over the year is 15 cm. The lake is in equilibrium with the underlying aquifer and has no inflow or outflow.
Solution: The mass balance equation is simple:
P = 50; % [cm/y]
Delta_S = 15; % [cm/y]
A = 1500; % [ha]
% Computing the annual evaporation
ET_volume = (P - Delta_S)*10^-2*A*10^4 % [m3/year]
ET_volume = 5250000
ET_flux = ET_volume/(365*A*10^4)*1000 % [mm/day]
ET_flux = 0.9589
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Example Problem 1.2: A city is supplied by water from a 1250-ha catchment area. The average water consumption of the community is 12,000 . The annual pre­cipitation in the region is 80 cm. A river with an average annual flow of 0.05 originates in the catchment and flows out of it. If the net annual groundwater outflow from the area is equivalent to a 5 cm depth of water per year, what is the evapotranspiration loss in cubic meters per year -- which, if exceeded, would cause a shortage of the water supply to the community? Assume the basin is in steady state condition in an annual scale.
Solution: The mass balance equation is simple:
P = 80; % [cm/year]
Q_out = 0.05; % [m3/s]
A = 1250; % [ha]
Q_consumption = 12000; % [m3/day]
G_out = 5; % [cm/year]
% Computing the annual evaporation
ET = P/100 - Q_out*24*3600*365/(A*10^4) - Q_consumption*365/(A*10^4) - G_out/100; % [m/year]
ET = ET*100 % [cm/year]
ET = 27.3456
Example Probelm 1.3: A golf course has requested a permit to install a 40,000 Square ft pond to enhance the beauty of its facility. Due to evaporation, the required water needs to be supplemented by groundwater pumping. There is a small creek that discharges 0.0005 m3/s into the pond. The outlet from the pond releases 0.003 m3/s of water to make sure there is a circulation in the pond. Precipitation rate is 260 mm/yr and the annual evaporation rate is 5 mm/day. The bottom of the pond is completely sealed from any seepage. The question is – how much groundwater pumping is needed to keep the pond in steady state condition?
Solution: The mass balance equation under steady state condition is as follows:
clear
P = 260; %[mm/yr]
E = 5; % [mm/day]
Q_in = 0.0005; % [m3/s]
Q_out = 0.003 % [m3/s]
Q_out = 0.0030
%-----
A = 40000*0.3048^2; % [m2]
G_in = (E*365/1000-P/1000)/A + Q_out - Q_in; %[m3/s]
G_in
G_in = 0.0029
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Example Problem 1.4: The following information is available for a lake including the water level above the sea level (feet), total annual precipitation (inches), total annual streamflow input (acre-feet) and annual lake evaporation (inches).
% Inputs of the problem
Years =["1";"2";"3";"4";"5"];
H= [4198.6;4197.7;4199.4;4203.4;4207.7]; % water level [ft]
P = [nan;9.46;16.78;17.43;28]; % Precip [inches]
Qin = [nan;1448900;2443000;5113390;6359170]; % [acre-feet]
E = [nan; 43.3;41.4;40.9;39.7]; % [inches]
table(Years,H,P,Qin,E)
ans = 5×5 table
 YearsHPQinE
1"1"4.1986e+03NaNNaNNaN
2"2"4.1977e+039.4600144890043.3000
3"3"4.1994e+0316.7800244300041.4000
4"4"4.2034e+0317.4300511339040.9000
5"5"4.2077e+0328635917039.7000
The following elevation-area-volume table for the lake is available. Is the lake shrinking or expanding during this five year period? Quntify your answer in terms of centemeter of water gain or loss. Note that the lake is terminal and has no outflow. Use avarega of the area at the begining and end of each calendar year for related computations.
% The lake Elevation-Area-Volume data
Lake_elev = [(4197:4210)';4212;4216;4218]; % [ft]
Lake_area = [839809;890047;969949;1079259;1140000;1175000;1201000;1223000;1250468;1330000;1375000;1410000;1450000;1490000;1572000;2228000;2519000]; %[ac]
Lake_volume = [12556430;13421890;14350140;15370180;16481450;17640700;18828700;20040700;21275600;22541900;23808300;25074700;26341000;27607300;30669000;38671000;43417000]; %[ac-ft]
table(Lake_elev,Lake_area,Lake_volume)
ans = 17×3 table
 Lake_elevLake_areaLake_volume
1419783980912556430
2419889004713421890
3419996994914350140
44200107925915370180
54201114000016481450
64202117500017640700
74203120100018828700
84204122300020040700
94205125046821275600
104206133000022541900
114207137500023808300
124208141000025074700
134209145000026341000
144210149000027607300
154212157200030669000
164216222800038671000
174218251900043417000
subplot 121
plot(Lake_volume,Lake_elev,'-o'); xlabel('Volume [acre-ft]'); ylabel('Elevation [feet]'); grid on; axis square
subplot 122
plot(Lake_volume,Lake_area,'-o'); xlabel('Area [acre]'); ylabel('Elevation [feet]'); grid on; axis square
Solution: The mass balance equation is simple: , where denotes the unaccounted inflows () or outflows ().
% Compute the average area for the first year
lake_area = @(h) interp1(Lake_elev,Lake_area,h); % compute lake area as a function of elevation [acre]
lake_vol = @(h) interp1(Lake_elev,Lake_volume,h); % compute lake volume as a function of elevation [acre-ft]
 
% Initialize the varaibles
A_ave = nan(5,1);
DS = nan(5,1);
net_loss = nan(5,1);
U = nan(5,1);
 
% Computing unaccounted water budgte for 2nd year
A_ave(2) = mean([lake_area(H(1)),lake_area(H(2))]); % average lake area
DS(2) = lake_vol(H(2)) - lake_vol(H(1)); % change of storage [acre-ft] : Volume at year end - Volume at year begining
net_loss(2) = A_ave(2)*(E(2)-P(2))/12; % A(E-P) -- Note the conversion of the unit from inches to feet
U(2) = DS(2)+ net_loss(2)-Qin(2); % U = DS+A(E-P)-Q [acre-ft]
 
% 3rd year
A_ave(3) = mean([lake_area(H(2)),lake_area(H(3))]); % average lake area
DS(3) = lake_vol(H(3)) - lake_vol(H(2)); % [acre-ft]
net_loss(3) = A_ave(3)*(E(3)-P(3))/12; % A(E-P)
U(3) = DS(3)+ net_loss(3)-Qin(3); % U = DS+A(E-P)-Q [acre-ft]
 
% 4th year
A_ave(4) = mean([lake_area(H(3)),lake_area(H(4))]); % average lake area
DS(4) = lake_vol(H(4)) - lake_vol(H(3)); % [acre-ft]
net_loss(4) = A_ave(4)*(E(4)-P(4))/12; % A(E-P)
U(4) = DS(4)+ net_loss(4)-Qin(4); % U = DS+A(E-P)-Q [acre-ft]
 
% 5th year
A_ave(5) = mean([lake_area(H(4)),lake_area(H(5))]); % average lake area
DS(5) = lake_vol(H(5)) - lake_vol(H(4)); % [acre-ft]
net_loss(5) = A_ave(5)*(E(5)-P(5))/12; % A(E-P)
U(5) = DS(5)+ net_loss(5)-Qin(5); % U = DS+A(E-P)-Q [acre-ft]
 
% showing the results in a table
table(Years,DS,net_loss,Qin,U)
ans = 5×5 table
 YearsDSnet_lossQinU
1"1"NaNNaNNaNNaN
2"2"-8.1659e+052.5563e+0614489002.9079e+05
3"3"1.5959e+061.9374e+0624430001.0903e+06
4"4"4.5553e+062.1744e+0651133901.6163e+06
5"5"5.3813e+061.2720e+0663591702.9414e+05
% --------------------------------------------------------------
% The above calculations can be placed in a "for" loop
% reinitialize the varaibles
A_ave = nan(5,1);
DS = nan(5,1);
net_loss = nan(5,1);
U = nan(5,1);
 
for i=2:5
A_ave(i) = mean([lake_area(H(i-1)),lake_area(H(i))]); % average lake area
DS(i) = lake_vol(H(i)) - lake_vol(H(i-1)); % change of storage [acre-ft] : Volume at year end - Volume at year begining
net_loss(i) = A_ave(i)*(E(i)-P(i))/12; % A(E-P) -- note the conversion of the unit from inches to feet
U(i) = DS(i)+ net_loss(i)-Qin(i); % U = DS+A(E-P)-Q [acre-ft]
end
U
U = 5×1
106 ×
NaN 0.2908 1.0903 1.6163 0.2941
It appears that over the last five years, the lake has had unaccounted inflows in all years.